Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

A person trying to lose weight by burning fat lifts a mass of $$10$$ $$kg$$ upto a height of $$1$$ $$m$$ $$1000$$ times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up? Fat supplies $$3.8 \times {10^7}J$$ of energy per $$kg$$ which is converted to mechanical energy with a $$20\% $$ efficiency rate. Take $$g = 9.8\,m{s^{ - 2}}$$ :

A

$$9.89 \times {10^{ - 3}}\,\,kg$$

B

$$12.89 \times {10^{ - 3}}\,kg$$

C

$$2.45 \times {10^{ - 3}}\,\,kg$$

D

$$6.45 \times {10^{ - 3}}\,\,kg$$

Assume the amount of fat is used = x kg

So total Mechanical energy available through fat

= $$x \times 3.8 \times {10^7} \times {{20} \over {100}}$$

And work done through lifting up

= 10 $$ \times $$ 9.8 $$ \times $$ 1000 = 98000 J

$$ \Rightarrow $$ $$x \times 3.8 \times {10^7} \times {{20} \over {100}}$$ = 98000

$$ \Rightarrow $$ $$x$$ = 12.89 $$ \times $$ 10^{-3} kg

So total Mechanical energy available through fat

= $$x \times 3.8 \times {10^7} \times {{20} \over {100}}$$

And work done through lifting up

= 10 $$ \times $$ 9.8 $$ \times $$ 1000 = 98000 J

$$ \Rightarrow $$ $$x \times 3.8 \times {10^7} \times {{20} \over {100}}$$ = 98000

$$ \Rightarrow $$ $$x$$ = 12.89 $$ \times $$ 10

2

MCQ (Single Correct Answer)

When a rubber-band is stretched by a distance $$x$$, it exerts restoring force of magnitude $$F = ax + b{x^2}$$ where $$a$$ and $$b$$ are constants. The work done in stretching the unstretched rubber-band by $$L$$ is :

A

$$a{L^2} + b{L^3}$$

B

$${1 \over 2}\left( {a{L^2} + b{L^3}} \right)$$

C

$${{a{L^2}} \over 2} + {{b{L^3}} \over 3}$$

D

$${1 \over 2}\left( {{{a{L^2}} \over 2} + {{b{L^3}} \over 3}} \right)$$

Given Restoring force, F = ax + bx^{2}

Work done in stretching the rubber-band by a distance $$dx$$ is

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dW = F\,dx = \left( {ax + b{x^2}} \right)dx$$

Intergrating both sides,

$$W = \int\limits_0^L {axdx + \int\limits_0^L {b{x^2}dx}}$$

= $$\left[ {a{{{x^2}} \over 2} + b{{{x^3}} \over 3}} \right]_0^L$$

= $${{a{L^2}} \over 2} + {{b{L^3}} \over 3}$$

Work done in stretching the rubber-band by a distance $$dx$$ is

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dW = F\,dx = \left( {ax + b{x^2}} \right)dx$$

Intergrating both sides,

$$W = \int\limits_0^L {axdx + \int\limits_0^L {b{x^2}dx}}$$

= $$\left[ {a{{{x^2}} \over 2} + b{{{x^3}} \over 3}} \right]_0^L$$

= $${{a{L^2}} \over 2} + {{b{L^3}} \over 3}$$

3

MCQ (Single Correct Answer)

This question has Statement $$1$$ and Statement $$2.$$ Of the four choices given after the Statements, choose the one that best describes the two Statements.

If two springs $${S_1}$$ and $${S_2}$$ of force constants $${k_1}$$ and $${k_2}$$, respectively, are stretched by the same force, it is found that more work is done on spring $${S_1}$$ than on spring $${S_2}$$.
**STATEMENT 1:** If stretched by the same amount work done on $${S_1}$$, Work done on $${S_1}$$ is more than $${S_2}$$
**STATEMENT 2:** $${k_1} < {k_2}$$

A

Statement 1 is false, Statement 2 is true

B

Statement 1 is true, Statement 2 is false

C

Statement 1 is true, Statement 2 is true, Statement 2 is the correct explanation for Statement 1

D

Statement 1 is true, Statement 2 is true, Statement 2 is not the correct explanation for Statement 1

We know force (F) = kx

$$W = {1 \over 2}k{x^2}$$

$$W =$$ $${{{{\left( {kx} \right)}^2}} \over {2k}}$$ $$\,\,\,$$

$$\therefore$$ $$W = {{{F^2}} \over {2k}}$$ [ as $$F=kx$$ ]

When force is same then,

$$W \propto {1 \over k}$$

Given that, $${W_1} > {W_2}$$

$$\therefore$$ $${k_1} < {k_2}$$

**Statement-2 is true.**

For the same extension, x_{1}
= x_{2}
= x

Work done on spring S_{1} is W_{1} = $${1 \over 2}{k_1}x_1^2 = {1 \over 2}{k_1}{x^2}$$

Work done on spring S_{2} is W_{2} = $${1 \over 2}{k_2}x_2^2 = {1 \over 2}{k_2}{x^2}$$

$$ \therefore $$ $${{{W_1}} \over {{W_2}}} = {{{k_1}} \over {{k_2}}}$$

As $${k_1} < {k_2}$$ then $${W_1} < {W_2}$$

**So, Statement-1 is false.**

$$W = {1 \over 2}k{x^2}$$

$$W =$$ $${{{{\left( {kx} \right)}^2}} \over {2k}}$$ $$\,\,\,$$

$$\therefore$$ $$W = {{{F^2}} \over {2k}}$$ [ as $$F=kx$$ ]

When force is same then,

$$W \propto {1 \over k}$$

Given that, $${W_1} > {W_2}$$

$$\therefore$$ $${k_1} < {k_2}$$

For the same extension, x

Work done on spring S

Work done on spring S

$$ \therefore $$ $${{{W_1}} \over {{W_2}}} = {{{k_1}} \over {{k_2}}}$$

As $${k_1} < {k_2}$$ then $${W_1} < {W_2}$$

4

MCQ (Single Correct Answer)

The potential energy function for the force between two atoms in a diatomic molecule is approximately given by $$U\left( x \right) = {a \over {{x^{12}}}} - {b \over {{x^6}}},$$ where $$a$$ and $$b$$ are constants and $$x$$ is the distance between the atoms. If the dissociation energy of the molecule is $$D = \left[ {U\left( {x = \infty } \right) - {U_{at\,\,equilibrium}}} \right],\,\,D$$ is

A

$${{{b^2}} \over {2a}}$$

B

$${{{b^2}} \over {12a}}$$

C

$${{{b^2}} \over {4a}}$$

D

$${{{b^2}} \over {6a}}$$

Given $$U\left( x \right) = {a \over {{x^{12}}}} - {b \over {{x^6}}}$$

$${U\left( {x = \infty } \right)}$$ = 0

We know $$F = - {{dU} \over {dx}} = - \left[ {{{12a} \over {{x^{13}}}} + {{6b} \over {{x^7}}}} \right]$$

At equilibrium: $${{dU\left( x \right)} \over {dx}} = 0$$

$$ \Rightarrow {{ - 12a} \over {{x^{13}}}} = {{ - 6b} \over {{x^7}}} $$

$$\Rightarrow x = {\left( {{{2a} \over h}} \right)^{{1 \over 6}}}$$

$$\therefore$$ $${U_{at\,\,equilibrium\,}} = {a \over {{{\left( {{{2a} \over b}} \right)}^2}}} - {b \over {\left( {{{2a} \over b}} \right)}}$$

$$ = - {{{b^2}} \over {4a}}$$

$$\therefore$$ $$D = 0 - \left( { - {{{b^2}} \over {4a}}} \right) = {{{b^2}} \over {4a}}$$

$${U\left( {x = \infty } \right)}$$ = 0

We know $$F = - {{dU} \over {dx}} = - \left[ {{{12a} \over {{x^{13}}}} + {{6b} \over {{x^7}}}} \right]$$

At equilibrium: $${{dU\left( x \right)} \over {dx}} = 0$$

$$ \Rightarrow {{ - 12a} \over {{x^{13}}}} = {{ - 6b} \over {{x^7}}} $$

$$\Rightarrow x = {\left( {{{2a} \over h}} \right)^{{1 \over 6}}}$$

$$\therefore$$ $${U_{at\,\,equilibrium\,}} = {a \over {{{\left( {{{2a} \over b}} \right)}^2}}} - {b \over {\left( {{{2a} \over b}} \right)}}$$

$$ = - {{{b^2}} \over {4a}}$$

$$\therefore$$ $$D = 0 - \left( { - {{{b^2}} \over {4a}}} \right) = {{{b^2}} \over {4a}}$$

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